Integrand size = 25, antiderivative size = 96 \[ \int e^{a+i b x} \cos (d+b x) \sin ^2(d+b x) \, dx=-\frac {i e^{a-i d-2 i (d+b x)}}{16 b}-\frac {i e^{a-i d+2 i (d+b x)}}{16 b}+\frac {i e^{a-i d+4 i (d+b x)}}{32 b}+\frac {1}{8} e^{a-i d} x \] Output:
-1/16*I*exp(a-I*d-2*I*(b*x+d))/b-1/16*I*exp(a-I*d+2*I*(b*x+d))/b+1/32*I*ex p(a-I*d+4*I*(b*x+d))/b+1/8*exp(a-I*d)*x
Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int e^{a+i b x} \cos (d+b x) \sin ^2(d+b x) \, dx=\frac {e^a \left (2 \left (\left (-i e^{2 i b x}+2 b x\right ) \cos (d)+\left (e^{2 i b x}-2 i b x\right ) \sin (d)\right )+e^{-2 i b x} \left (i \left (-2+e^{6 i b x}\right ) \cos (3 d)-\left (2+e^{6 i b x}\right ) \sin (3 d)\right )\right )}{32 b} \] Input:
Integrate[E^(a + I*b*x)*Cos[d + b*x]*Sin[d + b*x]^2,x]
Output:
(E^a*(2*(((-I)*E^((2*I)*b*x) + 2*b*x)*Cos[d] + (E^((2*I)*b*x) - (2*I)*b*x) *Sin[d]) + (I*(-2 + E^((6*I)*b*x))*Cos[3*d] - (2 + E^((6*I)*b*x))*Sin[3*d] )/E^((2*I)*b*x)))/(32*b)
Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {4972, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{a+i b x} \sin ^2(b x+d) \cos (b x+d) \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \left (\frac {1}{4} e^{a+i b x} \cos (b x+d)-\frac {1}{4} e^{a+i b x} \cos (3 b x+3 d)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i e^{a+2 i b x+i d}}{16 b}-\frac {3 e^{a+i b x} \sin (3 b x+3 d)}{32 b}-\frac {i e^{a+i b x} \cos (3 b x+3 d)}{32 b}+\frac {1}{8} x e^{a-i d}\) |
Input:
Int[E^(a + I*b*x)*Cos[d + b*x]*Sin[d + b*x]^2,x]
Output:
((-1/16*I)*E^(a + I*d + (2*I)*b*x))/b + (E^(a - I*d)*x)/8 - ((I/32)*E^(a + I*b*x)*Cos[3*d + 3*b*x])/b - (3*E^(a + I*b*x)*Sin[3*d + 3*b*x])/(32*b)
Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_ .) + (e_.)*(x_)]^(m_.), x_Symbol] :> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (74 ) = 148\).
Time = 0.84 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.86
method | result | size |
norman | \(\frac {\frac {3 \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{b}+\frac {x \,{\mathrm e}^{i b x +a}}{8}+\frac {x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{8}-\frac {x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{8}-\frac {x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}}{8}-\frac {i x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{4}-\frac {i x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{2}-\frac {i x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}}{4}+\frac {i {\mathrm e}^{i b x +a}}{8 b}-\frac {i {\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}}{8 b}+\frac {5 i {\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{8 b}-\frac {5 i {\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{8 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{3}}\) | \(275\) |
orering | \(-\frac {\left (-4 b x +i\right ) {\mathrm e}^{i b x +a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2}}{4 b}+\frac {i \left (b x +i\right ) \left (i b \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2}-{\mathrm e}^{i b x +a} b \sin \left (b x +d \right )^{3}+2 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} b \sin \left (b x +d \right )\right )}{4 b^{2}}-\frac {\left (-4 b x +i\right ) \left (-8 \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2} {\mathrm e}^{i b x +a} b^{2}-2 i b^{2} {\mathrm e}^{i b x +a} \sin \left (b x +d \right )^{3}+4 i \cos \left (b x +d \right )^{2} \sin \left (b x +d \right ) {\mathrm e}^{i b x +a} b^{2}+2 \cos \left (b x +d \right )^{3} {\mathrm e}^{i b x +a} b^{2}\right )}{16 b^{3}}+\frac {i x \left (10 \sin \left (b x +d \right )^{3} {\mathrm e}^{i b x +a} b^{3}-26 \cos \left (b x +d \right )^{2} \sin \left (b x +d \right ) {\mathrm e}^{i b x +a} b^{3}-22 i \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2} {\mathrm e}^{i b x +a} b^{3}+6 i \cos \left (b x +d \right )^{3} b^{3} {\mathrm e}^{i b x +a}\right )}{16 b^{3}}\) | \(336\) |
Input:
int(exp(a+I*b*x)*cos(b*x+d)*sin(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
(3/b*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^3+1/8*x*exp(a+I*b*x)+1/8*x*exp(a+I*b* x)*tan(1/2*b*x+1/2*d)^2-1/8*x*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^4-1/8*x*exp( a+I*b*x)*tan(1/2*b*x+1/2*d)^6-1/4*I*x*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)-1/2* I*x*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^3-1/4*I*x*exp(a+I*b*x)*tan(1/2*b*x+1/2 *d)^5+1/8*I/b*exp(a+I*b*x)-1/8*I/b*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^6+5/8*I /b*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^2-5/8*I/b*exp(a+I*b*x)*tan(1/2*b*x+1/2* d)^4)/(1+tan(1/2*b*x+1/2*d)^2)^3
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.64 \[ \int e^{a+i b x} \cos (d+b x) \sin ^2(d+b x) \, dx=\frac {{\left (4 \, b x e^{\left (2 i \, b x + a + i \, d\right )} + i \, e^{\left (6 i \, b x + a + 5 i \, d\right )} - 2 i \, e^{\left (4 i \, b x + a + 3 i \, d\right )} - 2 i \, e^{\left (a - i \, d\right )}\right )} e^{\left (-2 i \, b x - 2 i \, d\right )}}{32 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)*sin(b*x+d)^2,x, algorithm="fricas")
Output:
1/32*(4*b*x*e^(2*I*b*x + a + I*d) + I*e^(6*I*b*x + a + 5*I*d) - 2*I*e^(4*I *b*x + a + 3*I*d) - 2*I*e^(a - I*d))*e^(-2*I*b*x - 2*I*d)/b
Time = 0.20 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.61 \[ \int e^{a+i b x} \cos (d+b x) \sin ^2(d+b x) \, dx=\frac {x e^{a} e^{- i d}}{8} + \begin {cases} \frac {\left (256 i b^{2} e^{a} e^{6 i d} e^{4 i b x} - 512 i b^{2} e^{a} e^{4 i d} e^{2 i b x} - 512 i b^{2} e^{a} e^{- 2 i b x}\right ) e^{- 3 i d}}{8192 b^{3}} & \text {for}\: b^{3} e^{3 i d} \neq 0 \\x \left (\frac {\left (- e^{a} e^{6 i d} + e^{a} e^{4 i d} + e^{a} e^{2 i d} - e^{a}\right ) e^{- 3 i d}}{8} - \frac {e^{a} e^{- i d}}{8}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)*sin(b*x+d)**2,x)
Output:
x*exp(a)*exp(-I*d)/8 + Piecewise(((256*I*b**2*exp(a)*exp(6*I*d)*exp(4*I*b* x) - 512*I*b**2*exp(a)*exp(4*I*d)*exp(2*I*b*x) - 512*I*b**2*exp(a)*exp(-2* I*b*x))*exp(-3*I*d)/(8192*b**3), Ne(b**3*exp(3*I*d), 0)), (x*((-exp(a)*exp (6*I*d) + exp(a)*exp(4*I*d) + exp(a)*exp(2*I*d) - exp(a))*exp(-3*I*d)/8 - exp(a)*exp(-I*d)/8), True))
Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.01 \[ \int e^{a+i b x} \cos (d+b x) \sin ^2(d+b x) \, dx=\frac {4 \, {\left (b \cos \left (d\right ) e^{a} - i \, b e^{a} \sin \left (d\right )\right )} x + i \, \cos \left (4 \, b x + 3 \, d\right ) e^{a} - 2 i \, \cos \left (2 \, b x + 3 \, d\right ) e^{a} - 2 i \, \cos \left (2 \, b x + d\right ) e^{a} - e^{a} \sin \left (4 \, b x + 3 \, d\right ) - 2 \, e^{a} \sin \left (2 \, b x + 3 \, d\right ) + 2 \, e^{a} \sin \left (2 \, b x + d\right )}{32 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)*sin(b*x+d)^2,x, algorithm="maxima")
Output:
1/32*(4*(b*cos(d)*e^a - I*b*e^a*sin(d))*x + I*cos(4*b*x + 3*d)*e^a - 2*I*c os(2*b*x + 3*d)*e^a - 2*I*cos(2*b*x + d)*e^a - e^a*sin(4*b*x + 3*d) - 2*e^ a*sin(2*b*x + 3*d) + 2*e^a*sin(2*b*x + d))/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (55) = 110\).
Time = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.42 \[ \int e^{a+i b x} \cos (d+b x) \sin ^2(d+b x) \, dx=\frac {8 \, {\left (b x + d\right )} \cos \left (d\right ) e^{a} - 8 i \, {\left (b x + d\right )} e^{a} \sin \left (d\right ) + i \, {\left (e^{\left (4 i \, b x + 3 i \, d\right )} + e^{\left (-4 i \, b x - 3 i \, d\right )}\right )} e^{a} - 2 i \, {\left (e^{\left (2 i \, b x + 3 i \, d\right )} + e^{\left (-2 i \, b x - 3 i \, d\right )}\right )} e^{a} - 2 i \, {\left (e^{\left (2 i \, b x + i \, d\right )} + e^{\left (-2 i \, b x - i \, d\right )}\right )} e^{a} - 4 \, e^{a} \sin \left (2 \, b x + 3 \, d\right ) - 4 \, e^{a} \sin \left (-2 \, b x - d\right ) + 2 \, e^{a} \sin \left (-4 \, b x - 3 \, d\right )}{64 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)*sin(b*x+d)^2,x, algorithm="giac")
Output:
1/64*(8*(b*x + d)*cos(d)*e^a - 8*I*(b*x + d)*e^a*sin(d) + I*(e^(4*I*b*x + 3*I*d) + e^(-4*I*b*x - 3*I*d))*e^a - 2*I*(e^(2*I*b*x + 3*I*d) + e^(-2*I*b* x - 3*I*d))*e^a - 2*I*(e^(2*I*b*x + I*d) + e^(-2*I*b*x - I*d))*e^a - 4*e^a *sin(2*b*x + 3*d) - 4*e^a*sin(-2*b*x - d) + 2*e^a*sin(-4*b*x - 3*d))/b
Time = 15.02 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.17 \[ \int e^{a+i b x} \cos (d+b x) \sin ^2(d+b x) \, dx=\frac {x\,{\mathrm {e}}^a\,\left (\cos \left (d\right )-\sin \left (d\right )\,1{}\mathrm {i}\right )}{8}-\frac {{\mathrm {e}}^a\,\left (\cos \left (2\,b\,x\right )-\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )-\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,b}+\frac {{\mathrm {e}}^a\,\left (\cos \left (4\,b\,x\right )+\sin \left (4\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )+\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (2\,b\,x\right )+\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )+\sin \left (d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,b} \] Input:
int(cos(d + b*x)*exp(a + b*x*1i)*sin(d + b*x)^2,x)
Output:
(x*exp(a)*(cos(d) - sin(d)*1i))/8 - (exp(a)*(cos(2*b*x) - sin(2*b*x)*1i)*( cos(3*d) - sin(3*d)*1i)*1i)/(16*b) + (exp(a)*(cos(4*b*x) + sin(4*b*x)*1i)* (cos(3*d) + sin(3*d)*1i)*1i)/(32*b) - (exp(a)*(cos(2*b*x) + sin(2*b*x)*1i) *(cos(d) + sin(d)*1i)*1i)/(16*b)
Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.71 \[ \int e^{a+i b x} \cos (d+b x) \sin ^2(d+b x) \, dx=\frac {e^{b i x +a} \left (\cos \left (b x +d \right ) \sin \left (b x +d \right )^{2} i +\cos \left (b x +d \right ) b x +3 \sin \left (b x +d \right )^{3}-\sin \left (b x +d \right ) b i x -\sin \left (b x +d \right )\right )}{8 b} \] Input:
int(exp(a+I*b*x)*cos(b*x+d)*sin(b*x+d)^2,x)
Output:
(e**(a + b*i*x)*(cos(b*x + d)*sin(b*x + d)**2*i + cos(b*x + d)*b*x + 3*sin (b*x + d)**3 - sin(b*x + d)*b*i*x - sin(b*x + d)))/(8*b)